What is the question ? Show the first equality ? Treat the cases and .
Usually one would consider non-overlapping cases, but I guess the overlapping cases here would work.
Case 1: Suppose x y. What does this imply? If you don't recall what this implies I'll say that addition on R satisfy commutation, monotonicity, an equality for inverses, and an equality for neutrals. In other words, for all x, y, z, c on R the following properties for + hold:
If x y, (x+c) (y+c) (monotonicity)
Since you started with x y, do you see which property you want to use here first? Do you see what element (function of an element, depending on how you look at it) of R you want to select for "c"? If you do, I think you'll infer what I've hinted at. Then check your definition of absolute value. So in this case, abs(x-y) must equal... what? Then replace what abs(x-y) must equal in this case with abs(x-y) in the maximum formula.
Case 2: Suppose x>y. Now I'll point out that for all x, y, c in R,
If x>y, then (x+c)>(y+c). Now using a similar technique, along with your definition of absolute value you can show something about the absolute value in this case.
Then you might to want to indicate the cases exhaust R and the result consequently follows.
If the above doesn't give you enough clues, I'll happily try and spell out details here.